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Sagot :
To determine the final temperature of the gold nugget, we will use the principles of thermodynamics. Specifically, we will use the formula for heat transfer involving the change in temperature of a substance:
[tex]\[ Q = mc\Delta T \][/tex]
where:
- [tex]\( Q \)[/tex] is the heat lost or gained (in Joules),
- [tex]\( m \)[/tex] is the mass of the substance (in grams),
- [tex]\( c \)[/tex] is the specific heat capacity of the substance (in J/g°C),
- [tex]\( \Delta T \)[/tex] is the change in temperature (in °C).
Given the data:
- The mass of the gold nugget, [tex]\( m = 376 \text{ g} \)[/tex],
- The initial temperature, [tex]\( T_{\text{initial}} = 398 \text{ K} \)[/tex],
- The heat lost by the gold nugget, [tex]\( Q = 4850 \text{ J} \)[/tex] (since [tex]\( 4.85 \text{ kJ} = 4850 \text{ J} \)[/tex]),
- The specific heat capacity of gold, [tex]\( c = 0.128 \text{ J/g°C} \)[/tex].
First, we need to calculate the change in temperature ([tex]\( \Delta T \)[/tex]):
[tex]\[ \Delta T = \frac{Q}{mc} \][/tex]
Substituting the given values into the equation:
[tex]\[ \Delta T = \frac{4850 \text{ J}}{376 \text{ g} \times 0.128 \text{ J/g°C}} \][/tex]
Calculating the denominator:
[tex]\[ 376 \text{ g} \times 0.128 \text{ J/g°C} = 48.128 \text{ J/°C} \][/tex]
Now, dividing the heat lost by this product:
[tex]\[ \Delta T = \frac{4850 \text{ J}}{48.128 \text{ J/°C}} \approx 100.773 \text{ °C} \][/tex]
The initial temperature is given in Kelvin, and we need to find the final temperature in Kelvin as well. Thus, we need to ensure continuity in units.
The change in temperature, which is identical in Celsius and Kelvin for change only, [tex]\(\Delta T \approx 100.773 \text{ K} = 100.773 \text{ °C}\)[/tex].
Since heat is lost, the temperature of the gold nugget decreases. Thus, the final temperature ([tex]\( T_{\text{final}} \)[/tex]) is:
[tex]\[ T_{\text{final}} = T_{\text{initial}} - \Delta T \][/tex]
[tex]\[ T_{\text{final}} = 398 \text{ K} - 100.773 \text{ K} = 297.227 \text{ K} \][/tex]
Thus, the finalized and rounded answer for the final temperature of the gold nugget is:
[tex]\[ 297 \text{ K} \][/tex]
The correct answer is:
D) [tex]\( 297 \text{ K} \)[/tex]
[tex]\[ Q = mc\Delta T \][/tex]
where:
- [tex]\( Q \)[/tex] is the heat lost or gained (in Joules),
- [tex]\( m \)[/tex] is the mass of the substance (in grams),
- [tex]\( c \)[/tex] is the specific heat capacity of the substance (in J/g°C),
- [tex]\( \Delta T \)[/tex] is the change in temperature (in °C).
Given the data:
- The mass of the gold nugget, [tex]\( m = 376 \text{ g} \)[/tex],
- The initial temperature, [tex]\( T_{\text{initial}} = 398 \text{ K} \)[/tex],
- The heat lost by the gold nugget, [tex]\( Q = 4850 \text{ J} \)[/tex] (since [tex]\( 4.85 \text{ kJ} = 4850 \text{ J} \)[/tex]),
- The specific heat capacity of gold, [tex]\( c = 0.128 \text{ J/g°C} \)[/tex].
First, we need to calculate the change in temperature ([tex]\( \Delta T \)[/tex]):
[tex]\[ \Delta T = \frac{Q}{mc} \][/tex]
Substituting the given values into the equation:
[tex]\[ \Delta T = \frac{4850 \text{ J}}{376 \text{ g} \times 0.128 \text{ J/g°C}} \][/tex]
Calculating the denominator:
[tex]\[ 376 \text{ g} \times 0.128 \text{ J/g°C} = 48.128 \text{ J/°C} \][/tex]
Now, dividing the heat lost by this product:
[tex]\[ \Delta T = \frac{4850 \text{ J}}{48.128 \text{ J/°C}} \approx 100.773 \text{ °C} \][/tex]
The initial temperature is given in Kelvin, and we need to find the final temperature in Kelvin as well. Thus, we need to ensure continuity in units.
The change in temperature, which is identical in Celsius and Kelvin for change only, [tex]\(\Delta T \approx 100.773 \text{ K} = 100.773 \text{ °C}\)[/tex].
Since heat is lost, the temperature of the gold nugget decreases. Thus, the final temperature ([tex]\( T_{\text{final}} \)[/tex]) is:
[tex]\[ T_{\text{final}} = T_{\text{initial}} - \Delta T \][/tex]
[tex]\[ T_{\text{final}} = 398 \text{ K} - 100.773 \text{ K} = 297.227 \text{ K} \][/tex]
Thus, the finalized and rounded answer for the final temperature of the gold nugget is:
[tex]\[ 297 \text{ K} \][/tex]
The correct answer is:
D) [tex]\( 297 \text{ K} \)[/tex]
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